Question

A variable capacitor used in an RLC circuit produces a resonant frequency of $5.0\mathrm{MHz}$ when its capacitance is set to $15\mathrm{pF}$. What will the resonant frequency be when the capacitance is increased to $380\mathrm{pF}?$

Short answer

Answer: The final resonant frequency when the capacitance is increased to 380 pF is approximately 2.04 MHz.

Step-by-step solution

Identify the formula for resonant frequency

To find the resonant frequency in an RLC circuit, we will use the formula for the resonant frequency in RLC circuits: $f=\frac{1}{2\pi \sqrt{LC}}$ where $f$ is the resonant frequency, $L$ is the inductance, and $C$ is the capacitance.

Set up a proportion between initial and final conditions

Using the resonant frequency formula, we can set up a proportion between the initial and final conditions: $\frac{f_1}{f_2}=\frac{1}{2\pi \sqrt{L}}\cdot \frac{\sqrt{C_2}}{\sqrt{C_1}}$ where $f_1$ is the initial resonant frequency ($5.0\mathrm{MHz}$), $f_2$ is the final resonant frequency, $C_1$ is the initial capacitance ($15\mathrm{pF}$), and $C_2$ is the final capacitance ($380\mathrm{pF}$).

Solve for $f_2$

We can now solve for $f_2$: $f_2=f_1\cdot \frac{1}{2\pi \sqrt{L}}\cdot \frac{\sqrt{C_2}}{\sqrt{C_1}}\cdot \frac{2\pi \sqrt{L}}{1}$ $f_2=f_1\cdot \frac{\sqrt{C_2}}{\sqrt{C_1}}$ Substitute the given values: $f_2=5.0\mathrm{MHz}\cdot \frac{\sqrt{380\mathrm{pF}}}{\sqrt{15\mathrm{pF}}}$ $f_2=5.0\mathrm{MHz}\cdot \frac{\sqrt{380}}{\sqrt{15}}$ $f_2\approx 2.04\mathrm{MHz}$

State the final resonant frequency

Now we have found the final resonant frequency when the capacitance is increased to $380\mathrm{pF}$: $f_2\approx 2.04\mathrm{MHz}$